Calculating the infinite sum $1-\frac 1 7+\frac 1 9 - \frac{1}{15} + \frac 1 {17}\mp ...=\frac{1+\sqrt{2}}{8}\pi$

Question

Prove that $$1-\dfrac 1 7+\dfrac 1 9 - \dfrac{1}{15} + \dfrac 1 {17}\mp ...=\dfrac{1+\sqrt{2}}{8}\pi$$

My attempt: I tried to break it into two series $$(1+1/9+1/17+...)-(1/7+1/15+1/23+...)$$ But I don't know how to proceed. Any hints would be appreciated.

Answer 1

Using the hints by Mohammad Zuhair Khan and Feng Shao, let

$$f(x):=1-\sum_{n=0}^\infty\left(\frac{x^{8n-1}}{8n-1}-\frac{x^{8n+1}}{8n+1}\right).$$

Then if we differentiate term-wise,

$$f'(x)=-\sum_{n=0}^\infty(x^{8n-2}-x^{8n}).$$

Using the geometric sum formula,

$$f'(x)=-\frac{x^6}{1-x^8}+\frac{x^8}{1-x^8}=-\frac{x^6(1-x^2)}{1-x^8}.$$

Finally,

$$f(1)=1-\int_0^1\frac{x^6(1-x^2)}{1-x^8}dx.$$

https://www.wolframalpha.com/input/?i=integrate+x%5E6(1-x%5E2)%2F(1-x%5E8)+from+0+to+1

I see no easy way to solve the integral, other than by decomposition in simple fractions, which is tedious.

Answer 2

Just for your curiosity.

Since you received good hints and a good answer, let me show how we could compute the partial sum $$S_p=1- \sum_{n=1}^{p} \frac 1{8n-1}-\sum_{n=1}^{p}\frac 1{8n+1}$$ It write $$S_p=1+\frac{1}{8} \left(\psi \left(p+\frac{9}{8}\right)-\psi \left(p+\frac{7}{8}\right)-\psi \left(\frac{9}{8}\right)+\psi \left(\frac{7}{8}\right)\right)$$ where appears the digamma function.

Using the asymptotics and continuing with Taylor series for large values of $p$ $$S_p=\frac{ \pi}{8} \cot \left(\frac{\pi }{8}\right)+\frac{1}{32 p}-\frac{1}{64 p^2}+O\left(\frac{1}{p^3}\right)$$

Computing $$S_5=\frac{106748767459}{111928041225}\approx 0.953726$$ while the above truncated series would give $$\frac{\pi}{8} \cot \left(\frac{\pi }{8}\right)+\frac{9}{1600}\approx 0.953684$$

Just remember that, using the half angle, $\tan \left(\frac{\pi }{8}\right)=\sqrt 2 -1$ makes $\cot\left(\frac{\pi }{8}\right)=\sqrt 2 +1$

Answer 3

$$1-\dfrac 1 7+\dfrac 1 9 - \dfrac{1}{15} + \dfrac 1 {17}+...=1- \sum_{n=1}^{\infty} \frac 1{8n-1}-\frac 1{8n+1}$$ $$=1-\sum_{n=1}^{\infty} \frac 2{64n^2-1}=1-\frac{1}{32}\sum_{n=1}^{\infty} \frac 1{n^2-\frac{1}{8^2}}$$ and we have $$\frac{1-\pi x \cot(\pi x)}{2x^2}=\sum_{n=1}^{\infty }\frac{1}{n^2-x^2}$$ so $$1-\frac{1}{32}\sum_{n=1}^{\infty} \frac 1{n^2-\frac{1}{8^2}}=1-\frac{1}{32}\frac{1-\frac{\pi}{8}\cot(\pi/8)}{2(\frac{1}{64})}=\frac{1+\sqrt{2}}{8}\pi$$